'Get the parents from a json where string in javascript
I have this:
{ 'Payment' : {
'Referenced' : 'referenced payment',
'Conciliate' : 'conciliate payment',
'Multiple' : 'multiple payment'
}
}
but can change in all moment for random nodes or add more, like:
{ 'Payment' : {
'Referenced' : 'referenced payment',
'Conciliate' : 'conciliate payment',
'Multiple' : {
'mult1' : 'example1',
'mult2' : 'example1'
},
'Inventory' : {
'datastorage' : 'dt1'
}
}
All nodes can be asigned randomly, and I need to search by value, I can pass:
referenced payment
and need:
Payment/Referenced
or I send:
example1
and I need:
Payment/Multiple/mult1
I don't know if exist something like that.
Solution 1:[1]
// Function
const findPath = (obj, query) => {
const makeArray = (obj, path = []) => {
const pairs = Object.entries(obj);
return pairs.map(([key, value]) =>
typeof value === "object"
? makeArray(value, [...path, key])
: { path: [...path, key], value }
);
};
return (
makeArray(obj)
.flat(Infinity)
.find(({ path, value }) => value === query)?.path.join("/") ?? null
);
};
// Usage
const path1 = findPath(
{
Payment: {
Referenced: "referenced payment",
Conciliate: "conciliate payment",
Multiple: {
mult1: "example1",
mult2: {
test: 123,
},
},
Inventory: {
datastorage: "dt1",
},
},
},
123
);
const path2 = findPath(
{
Payment: {
Referenced: "referenced payment",
Conciliate: "conciliate payment",
Multiple: {
mult1: "example1",
},
Inventory: {
datastorage: "dt1",
},
},
},
"referenced payment"
);
console.log("123: " + path1);
console.log("referenced payment: " + path2);
Explanation
The first step is converting the object into a linear array of the object tree and its paths. This is done recursively. Then, the array is flattened in order to be iterated through with Array.prototype.find, if the value matches the query, the path is returned, if no match was found it returns null.
Credits
Thanks to @Bravo for suggesting path array instead of template literal
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
